Equation: $(eiaXf)=f(ax)$, where $X$ may be unbounded operator and $a in R>0$. I have found something but I am not convinced. The important point: Operator $X$ is not in terms of $a$ and the Hilbert space L^2(R>0, dx/x).
Hello Professor Mittal, yes, Operator is $X$ and $f$ is any vector in the Hilbert space L^2(R>0, dx/x). This operator $X$ is unknown. I am looking this operator $X$. This is one parameter unitary group so we can apply stones theorem to find $X$, since $X$ is self adjoint operator and may be bounded or unbounded, doesn't matter for the Stones theorem. But I can not success yet. Hi Professor Erkki, you can give me reference.
I enclose a paper, where I have carried out the proof of the generator of the scaling operation, see e.g. Eqs:V.3-V.6 in the enclosed Ann. Phys. article from 1983..
This particular generator concerns a theorem by Balslev and Combes, see ref. 37. and moreover this generator is an example of Stone's Theorem, M. H. STONE, The theory of representations for Boolean algebras, Transactions of the American Mathematical Society, vol. 40(1936), pp. 37-111.